1. 15.6 g of aluminum sulfide reacts with 10.0 g of water according to the reaction below: _Al₂S, (s) + H₂O (1)→ Al(OH), (aq) +_____ H₂S (g) a. Which substance is the limiting reactant? b. How many liters of H₂S is formed? c. What amount of excess reactant is left over, in grams?
1. 15.6 g of aluminum sulfide reacts with 10.0 g of water according to the reaction below: _Al₂S, (s) + H₂O (1)→ Al(OH), (aq) +_____ H₂S (g) a. Which substance is the limiting reactant? b. How many liters of H₂S is formed? c. What amount of excess reactant is left over, in grams?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Please solve all parts into one question. Thanks
![**How to Determine LR:**
1. Convert reactant quantities to moles.
2. Divide each mole amount by the balanced coefficient.
3. The smallest value is the limiting reactant (LR).
**Additional Information:**
- **Avogadro’s Number**: \(6.023 \times 10^{23}\) molecules/1 mole
- **STP Volume**: 1 mole = 22.4 L
- **Left Over**: Initial amount - Consumed amount
---
### Reaction Problem
15.6 g of aluminum sulfide reacts with 10.0 g of water according to the reaction below:
\[ \mathrm{Al_2S_3(s) + H_2O(l) \rightarrow Al(OH)_3(aq) + H_2S(g)} \]
**Questions:**
**a. Which substance is the limiting reactant?**
**b. How many liters of H₂S are formed?**
**c. What amount of excess reactant is left over, in grams?**
---
**Graph/Diagram Explanation**:
In the given image, there are no graphs or diagrams. The text consists of instructions and a chemical reaction problem meant to be solved by students.
Note to students:
- To solve for the limiting reactant, follow the three steps outlined above.
- Use Avogadro's number and STP volume as needed in your calculations.
- Make sure to balance the chemical equation before solving the problem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9359044e-f021-4fc6-a0ad-5a2dc758cf44%2F70806106-6a7c-4ee6-a018-fc215a9b00b1%2Fv9kipw_processed.png&w=3840&q=75)
Transcribed Image Text:**How to Determine LR:**
1. Convert reactant quantities to moles.
2. Divide each mole amount by the balanced coefficient.
3. The smallest value is the limiting reactant (LR).
**Additional Information:**
- **Avogadro’s Number**: \(6.023 \times 10^{23}\) molecules/1 mole
- **STP Volume**: 1 mole = 22.4 L
- **Left Over**: Initial amount - Consumed amount
---
### Reaction Problem
15.6 g of aluminum sulfide reacts with 10.0 g of water according to the reaction below:
\[ \mathrm{Al_2S_3(s) + H_2O(l) \rightarrow Al(OH)_3(aq) + H_2S(g)} \]
**Questions:**
**a. Which substance is the limiting reactant?**
**b. How many liters of H₂S are formed?**
**c. What amount of excess reactant is left over, in grams?**
---
**Graph/Diagram Explanation**:
In the given image, there are no graphs or diagrams. The text consists of instructions and a chemical reaction problem meant to be solved by students.
Note to students:
- To solve for the limiting reactant, follow the three steps outlined above.
- Use Avogadro's number and STP volume as needed in your calculations.
- Make sure to balance the chemical equation before solving the problem.
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